/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Graph each function. Then determ... [FREE SOLUTION] | 91Ó°ÊÓ

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Graph each function. Then determine any critical values, inflection points, intervals over which the function is increasing or decreasing, and the concavity. $$ g(x)=e^{-2 x} $$

Short Answer

Expert verified
The function \( g(x) = e^{-2x} \) is continuously decreasing and concave up everywhere with no critical points or inflection points.

Step by step solution

01

Understanding the Function

The function given is an exponential function, represented as \( g(x) = e^{-2x} \). This function has the base of the natural exponential function, \( e \), and is multiplied by \( -2x \), which results in a transformation involving exponential decay.
02

Graphing the Function

Recognize that \( e^{-2x} \) decreases as \( x \) increases because the exponent \( -2x \) is negative. The graph will be a decreasing curve starting from a y-intercept of 1 when \( x = 0 \) and approaching zero as \( x \) moves towards infinity. The curve is entirely in the first quadrant because \( e^{-2x} \) is always positive.
03

Finding the Critical Values

Critical values occur when the derivative is zero or undefined. Calculate the derivative: \( g'(x) = \frac{d}{dx}(e^{-2x}) = -2e^{-2x} \). The expression \( -2e^{-2x} \) is never zero or undefined (except at infinity), so there are no critical values.
04

Analyzing Intervals of Increase or Decrease

Since \( g'(x) = -2e^{-2x} < 0 \) for all \( x \), the function is decreasing for the entire domain of \( x \).
05

Identifying Inflection Points and Concavity

Find the second derivative to determine concavity: \( g''(x) = \frac{d}{dx}(-2e^{-2x}) = 4e^{-2x} \). Since \( g''(x) > 0 \) for all \( x \), the function is concave up everywhere, and there are no inflection points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Values
Critical values are important markers in the graph of a function where the slope is zero or where the derivative is undefined. These points help us identify where a function might change direction, such as moving from increasing to decreasing, or vice versa.
For the function \( g(x) = e^{-2x} \), we start by finding the first derivative, \( g'(x) = -2e^{-2x} \). Critical values appear where \( g'(x) = 0 \) or where the derivative does not exist. However, in this case, \( g'(x) = -2e^{-2x} \) will never equal zero or be undefined because the exponential function \( e^{-2x} \) is always positive, and thus \( e^{-2x} \) cannot equal zero.
As a result, the function \( g(x) \) has no critical values. This means the function will not change from increasing to decreasing (or vice versa) at any point. So, understanding there are no points where the slope is zero or undefined, allows us to predict the behavior of the graph throughout its domain without any surprises.
Inflection Points
Inflection points are points on the graph where the concavity changes, which can help in understanding the shape of the graph. These occur when the second derivative changes signs.
For our function \( g(x) = e^{-2x} \), we calculate the second derivative: \( g''(x) = 4e^{-2x} \). Since \( e^{-2x} \) is always positive, \( g''(x) \) is positive for all \( x \).
Because \( g''(x) \) does not change signs (it stays positive), there are no points where the concavity changes. Thus, the function \( g(x) = e^{-2x} \) has no inflection points. This consistent concavity helps confirm a particular smoothness to the curve across its domain.
Concavity
The concavity of a function indicates whether the graph is curving upwards or downwards. We determine concavity by evaluating the second derivative: if it is positive, the function is concave up; if negative, concave down.
For the exponential function \( g(x) = e^{-2x} \), we found that the second derivative is \( g''(x) = 4e^{-2x} \). This value is always positive since \( e^{-2x} \) remains a positive expression for all real numbers.
Therefore, the function is concave up everywhere on its domain. This means the graph of \( g(x) \) has a proper 'U' shape, with all parts of the graph bending upwards. This consistency in the graph's behavior provides a clear and predictable understanding of its shape across all values of \( x \).

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Most popular questions from this chapter

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